## Mixed boundary condition and uniqueness of solution

$\bf{Problem: }$ Let $L=a^{ij}D_{ij}u+b^i(x)D_iu+c(x)u$ is uniformly elliptic in a bounded domain $\Omega$, $c\leq 0$, $c/\lambda$ is bounded and $Lu=0$ in $\Omega$
(1) Let $\partial \Omega=S_1\cup S_2$ ($S_1$ non-empty) and assume an interior sphere condition at each point of $S_2$. Suppose $u\in C^2(\Omega)\cap C^1(\Omega\cup S_2)\cap C^0(\overline{\Omega})$ satisfies the mixed boundary condition
$\displaystyle u=0$ on $S_1$, $\displaystyle \sum \beta_iD_i u=0$ on $S_2$
where the vector $\boldsymbol{\beta}(x)=(\beta_1(x),\cdots,\beta_n(x))$ has a non-zero normal component(to the interior sphere) at each point $x\in S_2$, then $u\equiv 0$.

(2) Let $\partial \Omega$ satisfy an interior sphere condition, and assume that $u\in C^2(\Omega)\cap C^1(\overline{\Omega})$ satisfies the regular oblique derivative boundary condition
$\displaystyle \alpha(x)u+\sum \beta(x)D_iu=0$ on $\partial \omega$
where $\alpha(\vec{\beta}\cdot \vec{\nu}>;;;;;;;;;0$, $\nu$ is the outer normal vector. Then $u\equiv 0$.

$\mathbf{Proof: }$ (1) Suppose $u\not\equiv$ constant. If there exists $x\in \Omega$ such that $u(x)>;;;;;;;;0$, then $\exists\, x_0\in\partial S_2$ such that $\sup _{x\in \Omega}u=u(x_0)$. WLOG, assume $x_0$ is the origin and $\vec{\nu}$ is pointing to the negative $x_n-$axis. And $B$ is the interior ball at $x_0\in S_2$. Since $u$ is not constant, we know that $u(x)0$. This means $D_nu(x-0)>;0$
Since $u\in C^1(\Omega\cup S_2)$, then $u\in C^1(B\cup x_0)$. So $\displaystyle D_iu(x_0)=0$ for $i=1,2,\cdots,n-1$. So $\vec{beta}\cdot \vec{Du}=0$ means that $\beta_n=0$, this contradicts to the fact that $\vec{\beta}$ has non-zero component on $\vec{nu}$.
So $u\leq 0$ in $\Omega$. Similarly, $-u\leq 0$. So $u\equiv 0$.
(2) Use the same technique.

$\bf{Remark:}$ Gilbarg, Trudinger’ book. Chapter 3, exercise 3.1.